There is no standard way to deduce that a divisor $D$ on a projective variety $X$ is $m$-divisible in the Picard group, and even for $m=2$ this can be a difficult problem. Geometrically, this condition is equivalent to the existence of a simple cyclic cover $Y \to X$ of degree $m$ branched precisely on $D$.

The condition you give is not sufficient, and the following is a counterexample. Take a quartic surface $X \subset \mathbb{P}^3$ and assume that $X$ contains a conic $D$. Then the degree of $D$ with respect to the hyperplane section of $X$ is $2$. On the other hand, there is no divisor $L$ on $X$ such that $2L=D$. In fact, since $X$ is a $K3$ surface, by adjunction formula one finds $D \cdot D=-2$, that would imply $L \cdot L=-1/2$, a contradiction.

However, the condition is true for curves (see my comment below) and also in the following situation. Assume that $\textrm{Pic}(X)=\mathbb{Z}$, generated by an effective divisor $H$. If $\deg_H D = m \deg_H H$, then $D$ is linearly equivalent to $mH$. This happens for instance when $X=\mathbb{P}^n$, when $X \subset \mathbb{P}^3$ is a very general smooth surface of degree at least $4$ and when $X \subset \mathbb{P}^n$ is a smooth complete intersection of dimension at least $3$.